Answer:
I think it is right but not sure
Answer:

Step-by-step explanation:
−12 = 5[16] + b
80

If you want it in <em>Standard</em><em> </em><em>Form</em>:
y = 5x - 92
- 5x - 5x
__________

I am joyous to assist you anytime.
Answer:
The n th of the given sequence is 
Step-by-step explanation:
<u>Step 1</u> :-
Given sequence is 20,14,8,2,.......this sequence in arithmetic progression but this sequence is decreasing sequence.
given first term is 20 and difference is
now the nth term of given sequence is
by using formula 


final answer:-

<u>verification</u>:-

put n=1 we get first term is 20
put n=2 we get second term is 14
put n=3 we get third term is 8
put n=4 we get fourth term is 2
so the n th term of sequence is

One of the easier approaches would be that of x- and y-intercepts.
Setting x=0 results in y=-10; the y-intercept is (0, -10).
Setting y = 0 results in x = 5, so the x-intercept is (5,0).
Plot these 2 points. Then draw a str. line thru these points.
If you're using the app, try seeing this answer through your browser: brainly.com/question/2822258_______________
• Function: f(x) = 3x + 12.
A. Finding the inverse of f.
The composition of f with its inverse results in the identity function:
(f o g)(x) = x
f[ g(x) ] = x
3 · g(x) + 12 = x
3 · g(x) = x – 12
x – 12
g(x) = ⸺⸺
3
x g(x) = ⸺ – 4 <——— this is the inverse of f.
3________
B. Verifying that the composition of f and g gives us the identity function:
•

![\mathsf{=f\big[g(x)\big]}\\\\\\ \mathsf{=3\cdot \left(\dfrac{x}{3}-4\right)+12}\\\\\\ \mathsf{=\diagup\hspace{-7}3\cdot \dfrac{x}{\diagup\hspace{-7}3}-3\cdot 4+12}\\\\\\ \mathsf{=x-12+12}\\\\ \mathsf{=x\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7B%3Df%5Cbig%5Bg%28x%29%5Cbig%5D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%3D3%5Ccdot%20%5Cleft%28%5Cdfrac%7Bx%7D%7B3%7D-4%5Cright%29%2B12%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3D%5Cdiagup%5Chspace%7B-7%7D3%5Ccdot%20%5Cdfrac%7Bx%7D%7B%5Cdiagup%5Chspace%7B-7%7D3%7D-3%5Ccdot%204%2B12%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx-12%2B12%7D%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx%5Cqquad%5Cquad%5Ccheckmark%7D)
and also
•

![\mathsf{=g\big[f(x)\big]}\\\\\\ \mathsf{=\dfrac{f(x)}{3}-4}\\\\\\ \mathsf{=\dfrac{3x+12}{3}-4}\\\\\\ \mathsf{=\dfrac{\diagup\hspace{-7}3\cdot (x+4)}{\diagup\hspace{-7}3}-4}\\\\\\ \mathsf{=x+4-4}\\\\ \mathsf{=x\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7B%3Dg%5Cbig%5Bf%28x%29%5Cbig%5D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%3D%5Cdfrac%7Bf%28x%29%7D%7B3%7D-4%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%3D%5Cdfrac%7B3x%2B12%7D%7B3%7D-4%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3D%5Cdfrac%7B%5Cdiagup%5Chspace%7B-7%7D3%5Ccdot%20%28x%2B4%29%7D%7B%5Cdiagup%5Chspace%7B-7%7D3%7D-4%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx%2B4-4%7D%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx%5Cqquad%5Cquad%5Ccheckmark%7D)
________
C. Since f and g are inverse, then
f(g(– 2))
= (f o g)(– 2)
=
– 2 <span>✔
</span>
• Call h the compositon of f and g. So,
h(x) = (f o g)(x)
h(x) = x
As you can see above, there is no restriction for h. Therefore, the domain of h is R (all real numbers).
I hope this helps. =)